A continued fraction with a delayed repeating part

Question

I had previously seen how to solve for infinite continued fractions by solving a quadratic polynomial.

But in regard to the following I tried the same methods and could not seem to get it to work, maybe there is a good trick I could use?

$C=[3,6,1,4,1,4,1,4,.....]$

I tried to split it into different fractions and use reciprocals, but I cant seem to get any form that works for solving a quadratic or polynomial.

Answer 1

Let's ignore the $3,6$ for a sec. The continued fraction that remains gives rise to the equation

$$x=1 + \frac{1}{4+\frac{1}{x}}$$

This simplified to $4x^2-4x-1=0$. Apply the quadratic equation to get $x=\frac{1+\sqrt{2}}{2}$. Then plug this into

$$y=3+\frac{1}{6+\frac{1}{x}}$$ to get $y=\frac{1}{4}(14-\sqrt{2})$ which intriguingly has a decimal expansion begining $3.14$

Answer 2

Assume a repeating section of fraction as $x$.

$$x=1+\dfrac{1}{4+\dfrac{1}{x}}$$

Now you can find the value of $x$.

Then the original fraction is:

$$3+\dfrac{1}{6+\dfrac{1}{x}}$$